# Question #ab539

Jun 13, 2015

$R = 2.5 \cdot {10}^{4} N$.

#### Explanation:

The motion is a decelerated motion, so we can find the deceleration.

The two laws of the accelerated motion are:

$s = {s}_{0} + {v}_{0} t + \frac{1}{2} a {t}^{2}$

and

$v = {v}_{0} + a t$.

But if you solve the system of the two equations, you can find a third law really useful in those cases in which you haven't the time, or you haven't to find it.

${v}^{2} = {v}_{0}^{2} + 2 a \Delta s$ in which $\Delta s$ is the space run.

So:

$a = \frac{{v}^{2} - {v}_{0}^{2}}{2 \Delta s} = \frac{{0}^{2} - {200}^{2}}{2 \cdot 0.02} = - 1000000 \frac{m}{s} ^ 2 = - {10}^{6} \frac{m}{s} ^ 2$,

so the resistence would be:

$R = m \cdot a = 2.5 \cdot {10}^{-} 2 K g \cdot \left(- {10}^{6} \frac{m}{s} ^ 2\right) = - 2.5 \cdot {10}^{4} N$.